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Math Help - surjection of composition

  1. #1
    Member pberardi's Avatar
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    surjection of composition

    Below f is a function from a set A to a set B, and g is a function from the set B to a set C.

    Property 1: If f and g are surjections, then fg is a surjection.

    Proof of Property 1:
    Let z an arbitrary element in C.
    Then since f is a surjection, there is an element y in B such that z = f(y). Then since g is a surjection, there is an element x in A such that y = g(x). Hence by the definition of composite function, z = f(g(x)), that is z = fg(x). Hence fg is a surjection. QED

    I took this off of a website. I am wondering about its accuracy. Particularly the red line. Shouldn't it be something to the effect:

    Since g is a surjection, there is an element y in C such that z = g(y)? Aren't we talking about mapping all of C to A for g? I just don't see how A has anything to do with g. Please explain?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by pberardi View Post
    Below f is a function from a set A to a set B, and g is a function from the set B to a set C.

    Property 1: If f and g are surjections, then gf is a surjection.

    Proof of Property 1:
    Let z an arbitrary element in C.
    Then since g is a surjection, there is an element y in B such that z = g(y). Then since f is a surjection, there is an element x in A such that y = f(x). Hence by the definition of composite function, z = g(f(x)), that is z = gf(x). Hence gf is a surjection. QED

    I took this off of a website. I am wondering about its accuracy. Particularly the red line. Shouldn't it be something to the effect:

    Since g is a surjection, there is an element y in B such that z = g(y)?
    Hi pberardi.

    I have corrected some typos in your post. Perhaps it should be clearer now.
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